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1 O the Number of Crossig-Free Matchigs, (Cycles, ad Partitios Micha Sharir Emo Welzl Abstract We show that a set of poits i the plae has at most O(1005 perfect matchigs with crossig-free straight-lie embeddig The expected umber of perfect crossig-free matchigs of a set of poits draw iid from a arbitrary distributio i the plae is at most O(924 Seeral related bouds are deried: (a The umber of all (ot ecessarily perfect crossig-free matchigs is at most O(1043 (b The umber of left-right perfect crossig-free matchigs (where the poits are desigated as left or as right edpoits of the matchig edges is at most O(538 (c The umber of perfect crossig-free matchigs across a lie (where all the matchig edges must cross a fixed halig lie of the set is at most 4 These bouds are employed to ifer that a set of poits i the plae has at most O(8681 crossig-free spaig cycles (simple polygoizatios, ad at most O(1224 crossig-free partitios (partitios of the poit set, so that the coex hulls of the idiidual parts are pairwise disjoit 1 Itroductio Let P be a set of poits i the plae A geometric graph o P is a graph that has P as its ertex set ad its edges are draw as straight segmets coectig the correspodig pairs of poits The graph is crossigfree if o pair of its edges cross each other, ie, ay two edges are ot allowed to share ay poits other tha commo edpoits Therefore, these are plaar graphs with a plae embeddig gie by this specific drawig We are iterested i the umber of crossig-free geometric graphs o P of seeral special types Specifically, we cosider the umbers tr(p, of triagulatios (ie, maximal crossig-free graphs, pm(p, of crossig-free perfect matchigs, sc(p, of crossig-free spaig cycles, ad, cfp(p, of crossig-free partitios 1 (partitios Work by Micha Sharir has bee supported by the US-Israel Biatioal Sci Foudatio, by NSF Grat CCR , by a grat from the Israeli Acad of Sci for a Ceter of Excellece i Geom Comp at Tel Ai Ui, ad by the Herma Mikowski- MINERVA Ceter for Geometry at Tel Ai Ui School of Computer Sciece, Tel Ai Uiersity, Tel Ai 69978, Israel, ad Courat Ist of Math Sci, 251 Mercer Street, NYC, NY 10012, USA Ist Theoretische Iformatik, ETH Zürich, CH-8092 Zürich, Switzerlad 1 Our research was triggered by M a Kreeld askig about the umber of crossig-free partitios, ad, i the same week, by M Hoffma ad Y Okamoto askig about the umber of P, so that the coex hulls of the parts are pairwise disjoit We are cocered with upper bouds for the umbers listed aboe i terms of History This problem goes back to Newbor ad Moser [25] i 1980 who ask for the maximal possible umber of crossig-free spaig cycles i a set of poits 2 ; they gie a upper boud of ! but cojecture that the boud should be of the form c, c a costat This was established i 1982 by Ajtai, Chátal, Newbor, ad Szemerédi [4], who show 3 that there are at most crossig-free graphs 4 Further deelopmets were maily cocered with deriig progressiely better upper bouds for the umber of triagulatios 5 [29, 13, 28], so far culmiatig i a 59 upper boud by Satos ad Seidel [27] i 2003 It compares to Ω(848, the largest kow umber of triagulatios for a set of poits, recetly deried by Aichholzer et al [1]; this improes a earlier lower boud of 8 /poly( gie by García et al [17] (We let poly( deote a polyomial factor i Eery crossig-free graph is cotaied i some triagulatio (with at most 3 6 edges Hece, a c boud for the umber of triagulatios yields a boud of c < (8c for the umber of crossig-free graphs o a set of poits; with c 59, this is at most 472 To the best of our kowledge, all upper bouds so far o the umber of crossig-free graphs of arious types are deried ia a boud o the umber of triagulatios, of crossig-free spaig paths of a poit set (motiated by their quest for good fixed parameter algorithms for the plaar Euclidea Traelig Salesma Problem i the presece of a fixed umber of ier poits [10]; see also [19] 2 Akl s work [5] appeared earlier, but it refers to the mauscript by Newbor ad Moser, ad improes a lower boud (o the maximal umber of crossig-free spaig cycles of theirs 3 This paper is famous for its Crossig Lemma, proed i preparatio of the sigly expoetial boud The lemma gies a upper boud o the umber of edges a geometric graph with a gie umber of crossigs ca hae 4 For motiatio they metio also a questio of D Ais about the maximum umber of triagulatios a set of poits ca hae 5 Iterest was also motiated by the related questio (from geometric modelig [29] of how may bits it takes to ecode a triagulatio

2 Figure 1: 6 poits with 12 crossig-free perfect matchigs, the maximum possible umber; see [3] for the maximum umbers for up to te poits: 3 for 4 poits, 12 for 6, 56 for 8, ad 311 for 10 albeit i more refied ways Oe idea is to exploit the fact that graphs of certai types hae a fixed umber of edges; eg, sice a perfect matchig has 2 edges, we readily obtai pm(p ( 3 6 /2 tr(p < [14] A short historical accout of bouds o sc(p, with refereces icludig [5, 12, 17, 18, 20, 25, 26], ca be foud at the web site [11] (see also [8] The best boud published is 337 tr(p It relies o a 337 boud o the umber of cycles i a plaar graph [6] I the course of our iestigatios, we showed that a graph with m edges ad ertices has at most ( m cycles; hece, a plaar graph has at most 3 cycles The R Seidel proided us with a argumet, based o liear algebra, that a plaar graph has at most 6 < 245 spaig cycles Crossig-free partitios fit ito Figure 2: Graph of a crossig-free partitio the picture, sice eery such partitio ca be uiquely idetified with the graph of edges of the coex hulls of the idiidual parts these edges form a crossig-free geometric graph of at most edges; see Fig 2 The situatio is better uderstood for special cofiguratios, for example for P a set of poits i coex positio (the ertex set of a coex -go, where the Catala umbers C m := m+1( 1 2m m = Θ(m 3/2 4 m, m N 0, play a promiet role I coex positio tr(p = C 2 (the Euler-Seger problem, cf [30, pg 212] for its history, pm(p = C /2 for ee ([16], cf [30], sc(p = 1, ad cfp(p = C ([7] Crossig-free partitios for poit sets i coex positio costitute a well-established otio because of its may coectios to other problems, probably startig with plaar rhyme schemes i Becker s ote [7], cf [30, Solutio to 619pp] The geeral case was cosidered by [9] (uder the ame of pairwise liearly separable partitios for clusterig algorithms They show that that the umber of partitios ito k parts is O( 6k 12 for k costat Uder the assumptio of geeral positio (o three poits o a commo lie it is kow [17] that the umber of crossig-free perfect matchigs o a set of fixed size is miimized whe the set is i coex positio (Recetly, Aichholzer et al [1] showed that ay family of acyclic graphs has the miimal umber of crossig-free embeddigs o a poit set i coex positio With little surprise, the same holds for spaig cycles, but it does ot hold for triagulatios [21, 2, 23] For crossig-free partitios, this is ope Results We show the followig bouds, for a set P of poits i the plae: pm(p = O(1005, sc(p = O(8681, ad cfp(p = O(1224 Also, the expected umber of perfect crossig-free matchigs of a set of poits draw iid from ay distributio i the plae (where two radom poits coicide with probability 0 is O(924 The boud o the umber of crossig-free perfect matchigs is deried by a iductie techique that we hae adapted from the method that Satos ad Seidel [27] used for triagulatios (the adaptio howeer is far from obious We the go o to derie improed bouds o the umber of crossig-free matchigs of arious special types: (a The umber of all (ot ecessarily perfect crossig-free matchigs is at most O(1043 (b The umber of left-right perfect crossig-free matchigs (where the poits are desigated as left or as right edpoits of the matchig edges is at most O(538 (c The umber of perfect crossig-free matchigs across a lie (where all the matchig edges must cross a fixed halig lie of the set is at most 4 Fially, we derie upper bouds for the umbers of crossig-free spaig cycles ad crossig-free partitios of P i terms of the umber of certai types of matchigs of certai poit sets P that are costructed from P This yields the bouds as stated aboe We summarize the state of affairs i Table 1, (icludig lower bouds proofs are omitted here I work i progress, we are curretly refiig a tailored aalysis for spaig cycles ad trees, where the bouds ow stad at O(79 ad O(296, respectiely tr pm sc cfp P : 59 [27] P : 848 [1] 3 [17] 464 [17] 523 ma lrpm alpm rdpm P : P : Table 1: Etries c i the upper boud rows stad for O(c, ad etries c i the lower boud rows for Ω(c /poly(, where := P ma stads for all crossig-free matchigs, lrpm for perfect left-right crossig-free matchigs, alpm for perfect crossig-free matchigs across a lie, ad rdpm for the expected umber of perfect crossig-free matchigs of a set of iid poits

3 2 Matchigs: The Setup ad a Recurrece Let P be a set of poits i the plae i geeral positio, o three o a lie, o two o a ertical lie This is o costrait whe it comes to upper bouds o pm(p A crossig-free matchig M is a collectio of pairwise disjoit segmets whose edpoits belog to P Each poit of P is either matched, if it is a edpoit of a segmet of M, or isolated, otherwise The umber of matched poits is always ee If 2m poits are matched ad s poits are isolated, we call M a crossig-free m- matchig or (m, s-matchig We hae = 2m + s For m R we deote by ma m (P the umber of crossig-free matchigs of P with m segmets (this umber is 0 uless m {0, 1,, 2 }, ad by ma(p the umber of all crossig-free matchigs of P (ie ma(p = m ma m(p Recall pm(p = ma /2 (P Let M be a crossig-free (m, s-matchig o a set P of = 2m + s poits, as aboe The degree d(p of a poit p P i M is defied as follows It is 0 if p is isolated i M Otherwise, if p is a left (resp, right edpoit of a segmet of M, d(p is equal to the umber of isible left (resp, right edpoits of other segmets of M, plus the umber of isible isolated poits; isible meas ertically isible from the relatie iterior of the segmet of M that has p as a edpoit Thus p ad the other edpoit of the segmet are ot couted i d(p See Fig 3 Each left (resp, right ed- PSfrag replacemets poit u i M ca cotribute at u most 2 to the degrees of other poits: 1 to each of the left (resp, right edpoits of the segmets lyig ertically aboe ad below u, if there exist such z w Figure 3: Degrees i a matchig: d(u = 2, d( = 5, d(w = 1, d(z = 2 segmets Similarly, each isolated poit u ca cotribute at most 4 to the degrees of other poits: 1 to each of the edpoits of the segmets lyig ertically aboe ad below u It follows that p P d(p 4m + 4s There are may segmets ready for remoal The idea is to remoe segmets icidet to poits of low degree i a (m, s-matchig (poits of degree at most 3 or at most 4, to be specific We will show that there are may such poits at our disposal The, i the ext step, we show that segmets with a edpoit of low degree ca be reiserted i ot too may ways These two facts will be combied to derie a recurrece for the matchig cout For i N 0, let i = i (M deote the umber of matched poits of P with degree i i M Hece, i 0 i = 2m Lemma 21 Let, m, s N 0, with = 2m + s eery (m, s-matchig of ay set of poits, we hae ( s, ( s Proof Let P be the uderlyig poit set We hae i 0 i i = p P d(p 4s + 4m = 4s + i 0 2 i Therefore, 0 4s + i 0 (2 i i For κ R +, we add κ times = s + i 0 i to both sides to get κ (4 + κs + i 0 (2 + κ i i (4 + κs + 0 i<2+κ (2 + κ i i We set κ = 2 for (21 ad κ = 3 for (22 There are ot too may ways of isertig a segmet Fix some p P ad let M be a crossigfree matchig which leaes p isolated Now we match p with some other isolated poit such that the oerall matchig cotiues to be crossig-free For i N 0, let h i = h i (p, P, M be the umber of ways that ca be doe so that p has degree i after its isertio Lemma 22 (23 (24 4h 0 + 3h 1 + 2h 2 + h 3 24, 5h 0 + 4h 1 + 3h 2 + 2h 3 + h 4 48 Proof Let l i = l i (p, P, M be the umber of ways we ca match the poit p as a left edpoit of degree i First, we claim that l 0 {0, 1} To show this, form the ertical decompositio of M by drawig a ertical segmet up ad dow from each (matched or isolated poit of P \{p}, ad exted these segmets util they meet a edge of M, or else, all the way to ifiity; see Fig 4 We call these ertical segmets walls i order to distiguish them from the segmets i the matchig We obtai a decompositio of the plae itopsfrag ertical replacemets trapezoids Let τ be the trapezoid cotaiig p (assumig geeral u positio, p lies i the iterior of τ See Fig 4 We moe from τ to the right through ertical walls to adjacet trapezoids util we reach a ertical wall that is determied τ p I Figure 4: Isertig a segmet at p; d(p = 1 after isertio by a poit that is either a left edpoit or a isolated poit (if at all we may make our way to ifiity whe p caot be matched as a left edpoit to ay poit, i which case l i = 0 for all i Note that up to that poit there was always a uique choice for the ext trapezoid to eter Eery crossig-free segmet with p as its left edpoit will hae to go through all of these trapezoids It coects either to (which ca happe oly if is isolated,

4 or crosses the ertical wall up or dow from The former case yields a segmet that gies p degree 0 I the latter case, will cotribute 1 to the degree of p So p, if a optio, is the oly possible segmet that lets p hae degree 0 as a left edpoit (p will ot be a optio whe it crosses some segmet, or whe is a left edpoit We will retur to this set-up whe we cosider degrees 1, i which case acts as a bifurcatio poit Before doig so, we first itroduce a fuctio f It maps eery oegatie real ector (λ 0, λ 1,, λ k of arbitrary legth k + 1 N to the maximum possible alue the expressio (25 λ 0 l 0 + λ 1 l λ k l k ca attai (for ay isolated poit i ay matchig of ay fiite poit set of ay size We hae already show that f(λ λ for λ R + 0 We claim that for all (λ 0, λ 1,, λ k (R + 0 k+1, with k 1, we hae { λ0 + f(λ (26 f(λ 0, λ 1,, λ k max 1,, λ k, 2f(λ 1,, λ k Assumig (26 has bee established, we ca coclude that f(1 1, f(2, 1 3, f(3, 2, 1 6, ad f(4, 3, 2, 1 12; that is, 4l 0 + 3l 1 + 2l 2 + l 3 12 ad the first iequality of the lemma follows, sice the same boud clearly holds for the case whe p is a right edpoit The secod iequality follows similarly from f(5, 4, 3, 2, 1 24 It remais to proe (26 Cosider a costellatio with a poit p that realizes the alue of f(λ 0, λ 1,, λ k We retur to the set-up from aboe, where we hae traced a uique sequece of trapezoids from p to the right, till we ecoutered the first bifurcatio poit (if does ot exist the all l i aish Case 1: is isolated We kow that λ 0 l 0 λ 0 If we remoe from the poit set, the eery possible crossig-free segmet emaatig from p to its right has its degree decreased by 1 Therefore, λ 1 l λ k l k f(λ 1,, λ k, so the expressio (25 caot exceed λ 0 + f(λ 1,, λ k i this case Case 2: is a matched left edpoit The λ 0 l 0 = 0 (that is, we caot coect p to Possible crossigfree segmets with p as a left edpoit are discrimiated accordig to whether they pass aboe or below We first cocetrate o the segmets that pass aboe ; we call them releat segmets (emaatig from p Let l i be the umber of releat segmets that gie p degree i We carefully remoe isolated poits from P \ {p} ad segmets with their edpoits from the matchig M (eetually also the segmet of which is a left edpoit, so that i the ed all releat segmets are still aailable ad each oe, if iserted, makes the degree of p exactly 1 uit smaller tha its origial alue (this deletio process may create ew possibilities for segmets from p That will show λ 1 l λ k l k f(λ 1,, λ k The same will apply to segmets that pass below, usig a symmetric argumet, which gies the boud of 2f(λ 1,, λ k for (25 i this secod case The remoal process is performed as follows We defie a relatio o the set whose elemets are the edges of M ad the sigleto sets formed by the isolated poits of P \ {p}: a b if a poit a a is ertically isible from a poit b b, with a below b As is well kow (cf [15, Lemma 114], is acyclic Let + deote the trasitie closure of, ad let deote the trasitie reflexie closure of Let e be the segmet with as its left edpoit, ad cosider a miimal elemet a with a + e Such a elemet exists, uless e itself is a miimal elemet with respect to a is a sigleto: So it cosists of a isolated poit; with abuse of otatio we also deote by a the isolated poit itself a caot be a poit to which p ca coect with a releat edge Ideed, if this were the case, we add that edge e = pa ad modify to iclude e too; more precisely, ay pair i that ioles a is replaced by a correspodig pair iolig e, ad ew pairs iolig e are added (clearly, the relatio remais acyclic ad all pairs related uder + cotiue to be so related after e is icluded ad replaces a See Fig 5(a We hae e e (sice, by assumptio, the left edpoit of e is ertically isible below e, ad e + e (sice the right edpoit a of e satisfies a + e a cotradictio With a similar reasoig we ca rule out the possibility that a cotributes to the degree of p whe matched ia a releat edge pq Ideed, if this were the case, let e be the segmet directly aboe a, which is the first lik i the chai that gies a + e, ie, a e e (e must exist sice a + e After addig pq with a cotributig to its degree, we hae either a pq ad pq e (see Fig 5(b, or we hae pq a (see Fig 5(c I the former case, we hae a pq e e pq cotradictig the acyclicity of I the latter case, we hae pq a + e pq, agai a cotradictio So if we remoe a, the all releat edges from p remai i the game ad the degree of each of them (ie, the degree of p that the edge iduces whe iserted does ot chage p e e a p e e (a (b (c Figure 5: (a The poit a caot be coected to p ia a releat edge (b,c a caot cotribute from below (i (b or from aboe (i (c to the degree of p whe a releat edge pq is iserted a q p e e a q

5 a is a edge: It caot obstruct ay isolated poit or left edpoit below it from cotributig to the degree of a releat edge pq aboe (because a is miimal with respect to If a obstructs a cotributio to a releat edge pq from aboe, the we add pq, thus pq a which, together with PSfrag a replacemets + e ad e pq, cotradicts the acyclicity of (Fig 6 Agai, we ca remoe a without ay chages to releat possible edges from p We keep successiely re- a moig elemets util e is miimal with respect to Note p q e that so far all the releat edges from p are still possible, ad the degree of p that ay of them iduces whe iserted has ot Figure 6: Edge a caot obstruct a poit from cotributig from aboe to the degree of p whe a releat edge pq is iserted chaged Now we remoe e with its edpoits This caot clear the way for ay ew cotributio to the degree of a releat edge I fact, ay such degree decreases by exactly 1 because disappears The claim is show, ad the proof of the lemma is completed Deriig a recurrece Lemma 23 Let, m N 0, such that m 2 ad s := 2m For eery set P of poits, we hae 12(s+2 3s ma m 1 (P if s < 3 ma m (P, ad 16(s+2 7s/3 ma m 1(P if s < 3 7 Let us ote right away that the first iequality supersedes the secod for s < 5 (ie m > 2 5, while the secod oe is superior for s > 5 Proof Fix the set P, ad let X ad Y be the sets of all crossig-free m-matchigs ad (m 1-matchigs, respectiely, i P Let us cocetrate o the first iequality We defie a edge-labeled bipartite graph G o X Y as follows: Gie a m-matchig M, if p is a edpoit of a segmet e M ad d(p 3, the we coect M X to the (m 1-matchig M \ {e} Y with a edge labeled (p, d(p; d(p is the degree label of the edge Note that M ad M \ {e} ca be coected by two (differetly labeled edges, if both edpoits of e hae degree at most 3 For 0 i 3, let x i deote the umber of edges i G with degree label i We hae (2 6s X 4x 0 + 3x 1 + 2x 2 + x 3 24(s + 2 Y }{{} }{{} ma m(p ma m 1(P The first iequality is a cosequece of iequality (21 of Lemma 21 The secod iequality is implied by iequality (23 i Lemma 22, as follows For a fixed (m 1-matchig M i P, cosider a edge of G that is icidet to M ad is labeled by (p, i (if there is such a edge The p must be oe of the s+2 isolated poits of P (with respect to M, ad there is a way to coect p to aother isolated poit i a crossig-free maer, so that p has degree i i the ew matchig Hece, the cotributio by p ad M to the sum 4x 0 +3x 1 +2x 2 +x 3 is at most 24, by iequality (23 i Lemma 22, ad the right iequality follows The combiatio of both iequalities yields the first iequality the lemma By cosiderig edpoits up to degree 4 (istead of 3, we get the secod iequality For m, N 0, let ma m ( be the maximum umber of crossig-free m-matchigs a set of poits ca hae Lemma 24 For s, m, N 0, with = 2m + s, ma 0 (0 = 1, s ma m( 1, for s 1, 12(s+2 ma m ( 3s ma m 1 (, for s < 3, 16(s+2 7s/3 ma m 1(, for s < 3 7 Proof ma 0 (0 = 1 is triial The first of the three iequalities is implied by s ma m (P = p P ma m(p \ {p} ma m ( 1, for ay set P of poits The secod ad third iequalities follow from Lemma 23 3 Solig a Recurrece We derie a upper boud for a fuctio G G λ,µ : N 2 0 R +, for a pair of parameters λ, µ R +, µ 1, which satisfies (with s := 2m G(0, 0 = 1, { s G(m, 1, for s 1, (37 G(m, λ(s+2 µs G(m 1,, for s < µ The recurrece i Lemma 24 implies that a upper boud o G 12,3 (m, seres also as a upper boud for ma m (, ad the same holds for G 16,7/3 (m, We will see how to best combie the two parameter pairs, to obtai ee better bouds for ma m ( Later, we will ecouter other istaces of this recurrece, with other alues of λ ad µ We diide by λ m µ m The (37 becomes G(m, G(m, 1 λ m µ m µs λ m µ, for s 1, 1 m µ(s+2 µs G(m 1, λ m 1 µ m+1, for s < µ G(m, We set H(m, = H µ (m, := λ m µ Therefore, m still with the coetio s := 2m ad the assumptio µ 1, we hae (ote idepedece of λ H(0, 0 = 1, (38 H(m, { µs H(m, 1, for s 1, µ(s+2 µs H(m 1,, for s < µ

8 ( N 2m ( N (4λ(µ 1 m 2m r 2m ( N poly(m r m (4λ(µ 1 m( ( r 2m = m ( r 2m We see that if we sample r poits from a large eough set, the the expected umber of crossig-free matchigs obseres for all m the upper boud deried for the rage of small m Suppose ow that, for ee, we sample iid poits from a arbitrary distributio, for which we oly require that two sampled poits coicide with probability 0 The we ca first sample a set P of N > 11 8 poits, ad the choose a subset of size uiformly at radom from the family of all subsets of this size We obtai a set R of iid poits from the gie distributio If P is i geeral positio, by the argumet aboe the expected umber of perfect crossig-free matchigs is at most ( /2 If P exhibits colliearities, we perform a small perturbatio yieldig a set P ad the subset R Now the boud applies to R, ad also to R sice a sufficietly small perturbatio caot decrease the umber of crossigfree perfect matchigs Theorem 42 For ay distributio i the plae for which two sampled poits coicide with probability 0, the expected umber of crossig-free perfect matchigs of iid poits is at most ( /2 poly( = O( Left-Right Perfect Matchigs Here we assume that P is partitioed ito two disjoit subsets L, R ad cosider bipartite matchigs i L R such that, for each edge of the matchig, its left edpoit belogs to L ad its right edpoit to R We modify the defiitio of the degrees of the poits: If p L is a matched to a poit i R, the d(p is equal to the umber of left edpoits plus the umber of right-labeled isolated poits that are ertically isible from (the relatie iterior of e A symmetric defiitio holds for right edpoits (Ituitiely, a rightlabeled isolated poit q has to cotribute oly to the degrees of left-labeled poits, because, whe we isert a right edpoit, it caot coect to q, ad it does ot matter whether its icidet edge passes aboe or below q; that is, q does ot cause ay bifurcatio i the ways i which p ca be coected Sice isolated poits cotribute ow oly 2 to degrees of edpoits, we hae p P d(p 4m + 2s The aalysis further improes, because whe we reisert a poit p L, say, the correspodig umbers h i must be equal to l i, sice p ca oly be the left edpoit of a matchig edge A similar improemet holds for poits q R Hece, we ca boud the sum 4h 0 + 3h 1 + 2h 2 + h 3 by 12, rather tha 24; similarly, we hae 5h 0 + 4h 1 + 3h 2 + 2h 3 + h 4 24 That is, we hae for (λ, µ the pairs (6, 2 ad (8, 5 3 aailable ( k 1 i=0 6(2i+2 4i We ifer a boud of G 8,5/3 ( 2 k,, for k = 6, implyig Theorem 43 Let P be a set of poits i the plae ad assume that the poits are classified as left edpoits or right edpoits The umber of leftright ( perfect crossig-free matchigs i P is at most 2 7/10 3 3/20 7 7/10 poly( = O( Matchigs Across a Lie Cosider ext the special case of crossig-free bipartite perfect matchigs betwee two sets of 2 poits each that are separated by a lie Here we ca obtai a upper boud that is smaller tha the oe i Theorem 43 Theorem 44 Let be a ee iteger The umber of crossig-free perfect bipartite matchigs betwee two separated sets of 2 poits each i the plae is at most C 2 /2 < 4 ; (C m is the mth Catala umber Proof Let L ad R be the gie separated sets Without loss of geerality, take the separatig lie λ to be the y-axis, ad assume that the poits of L lie to the left of λ ad the poits of R lie to its right Let M be a crossig-free perfect bipartite matchig i L R For each edge e of M, let e L (resp, e R deote the portio of e to the left (resp, right of λ, ad refer to them as the left half-edge ad the right half-edge of e, respectiely We will obtai a upper boud for the umber of combiatorially differet ways to draw the left half-edges of a crossig-free perfect matchig i L R The same boud will apply symmetrically to the right half-edges, ad the fial boud will be the square of this boud I more detail, we igore R, ad cosider collectios S of 2 pairwise disjoit segmets, each coectig a poit of L to some poit o λ, so that each poit of L is icidet to exactly oe segmet For each segmet i S, we label its λ-edpoit by the poit of L to which it is coected The icreasig y-order of the λ-edpoits of the segmets thus defies a permutatio of L, ad our goal is to boud the umber of differet permutatios that ca be geerated i this way (I geeral, this is a strict upper boud o the quatity we seek We obtai this boud i the followig recursie maer Write m := L = 2 Sort the poits of L from left to right (we may assume that there are o ties they ca be elimiated by a slight rotatio of λ, ad let p 1, p 2,, p m deote the poits i this order Cosider the half-edge e 1 emaatig from the leftmost poit p 1 Ay other poit p j lies either aboe or below

9 e 1 By rotatig e 1 about p 1, we see that there are at most m (exactly m, if we assume geeral positio ways to split {p 2,, p m } ito a subset L + 1 of poits that lie aboe e 1 ad a complemetary subset L 1 of poits that lie below e 1, where i the i-th split, L + 1 = i 1 ad L 1 = m i Note that, i ay crossig-free perfect bipartite matchig that has e 1 as a left halfedge icidet to p 1, all the poits of L + 1 (resp, of L 1 must be icidet to half-edges that termiate o λ aboe (resp, below the λ-edpoit of e 1 Hece, after haig fixed i, we ca proceed to boud recursiely ad separately the umber of permutatios iduced by L + 1, ad the umber of those iduced by L 1 I other words, deotig by Π(m the maximum possible umber of differet permutatios iduced i this way by a set L of m poits (i geeral positio, we get the recurrece Π(m m i=1 Π(i 1Π(m i, for m 1, where Π(0 = 1 Howeer, this is the recurrece that (with equality defies the Catala umbers, so we coclude that Π(m C m A (probably weak upper boud for the umber of crossig-free perfect bipartite matchigs i L R is thus C 2 m Ideed, for ay permutatio π L of L ad ay permutatio π R of R, there is at most oe crossig-free perfect bipartite matchig i L R that iduces both permutatios Namely, it is the matchig that coects the j-th poit i π L to the j-th poit i π R, for each j = 1,, m The asserted boud of C 2 m =C 2 /2 <4 follows 5 Two Implicatios Spaig Cycles Theorem 51 Let P be a set of poits i the plae The the umber of crossig-free spaig cycles satisfies sc(p (2 7/5 3 7/10 7 7/5 poly( = O( Proof We costruct (from P a ew set P of 2 poits by creatig two copies p +, p of each p P, ad by placig these copies co-ertically close to the origial locatio of p, with p + aboe p Let π be a cycle i P We map π to a perfect matchig i P : For each p P, let q, r be its eighbors i π (i If both q, r lie to the left of p, with the edge qp lyig aboe rp, we coect p + to either q + or q, ad coect p to either r + or r (the actual choices will be determied at q ad r by similar rules (ii The same rule applies whe both q, r lie to the right of p (iii If q lies to the left of p ad r lie to the right of p, the we coect p + to either q + or q, ad coect p to either r + or r Clearly, the resultig graph π is a crossig-free perfect matchig i P, assumig geeral positio of the poits of P, if we draw each pair p +, p sufficietly close to each other We assig to each poit p P a label that depeds o π A poit whose two eighbors i π lie to its left is labeled as a right poit, a poit whose two eighbors i π lie to its right is labeled as a left poit, ad a poit haig oe eighbor i π to its right ad oe to its left is labeled as a middle poit We assig the cycle π to the pair (π, λ, where π is the resultig perfect matchig o P ad λ is the labelig of P, as just defied Each pair (π, λ ca be realized by at most oe cycle π i P, by mergig each pair p +, p back ito the origial poit p (i geeral, the resultig graph is a collectio of pairwise disjoit cycles It therefore suffices to boud the umber of such pairs (π, λ A gie labelig λ of P uiquely classifies each poit of P as beig either a left poit of a edge of the matchig or a right edpoit of such a edge Hece, the umber of crossig-free perfect matchigs π o P that respect this left-right assigmet is at most (2 7/10 3 3/20 7 7/10 2 poly( The umber of labeligs of P is 3 Hece, the umber of crossig-free cycles i P is at most (2 7/5 3 7/10 7 7/5 poly(, as asserted Clearly, it follows from the proof that the boud holds for the umber of crossig-free spaig paths as well, ad also for the umber of cycle coers (or path coers of P Crossig-free Partitios For a boud o cfp(p, we relate crossig-free partitios of a poit set P to matchigs To this ed, eery crossig-free partitio is mapped to a tuple (M, S, I +, I where (see Fig 7 (i M is the matchig i P, whose edges coect the leftmost to the rightmost poit of each set with at least two elemets (such a segmet is called the spie of its set, (ii S is the set of all poits that form sigleto sets i the partitio, ad (iii I + (resp, I is the set of poits i P \ S that are either leftmost or the rightmost i their set, ad which lie aboe (resp, below the spie of their set M is crossig-free, ad the partitio is uiquely determied by (M, S, I +, I Therefore, ay boud o the umber of such tuples will establish a boud o the umber of crossig-free partitios For eery crossig-free matchig M o P there are 3 2 M triples (S, I +, I which form a 4- tuple with M (ot all of them hae to come from a crossig-free partitio Therefore m 3 2m ma m (P is a boud o the umber of crossig-free partitios Igorig the 3 -factor for the time beig, we hae to determie a upper boud o 3 2m ma m (P, for which we employ the boud from (42 We obsere that 3 2m G λ,µ (m, = G λ/9,µ (m,, ad therefore G 16/9,7/3 (m,, m 2 3 2m 5, ma m (P G 4/3,3 (m, G 16,7/3( 2 5,, otherwise G 12,3( 2 5,

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